Proof Using that L is $\Delta ^0_2$ we can approximate L in stages. We keep track of the blocks that $\emptyset ^{\prime }_s$ thinks are in $L_s$ and build corresponding blocks in $D_s$ . When we see two blocks in $L_s$ change order or merge, we keep the representative of the block with the smallest member (in the sense of $<_{\mathbb {N}}$ ) and remove the other one by densifying (i.e., adding points so that the block becomes part of a copy of $\mathbb {Q}$ ). Then we add a new block in the correct place if needed.

More formally, let $(L_s,<_s,B_s){}_{s}$ be a sequence of linear orders with block relation that has limit $(L,<_L, B)$ where each $L_s\subseteq L_{s+1}$ is a subset of $\omega $ .

Define $D_0=\emptyset $ . We will keep a follower function $f_s$ from the blocks of $D_s$ to a corresponding element in $L_s$ that represents the block. At stage s, for any $b_i,b_j\in \mathrm {dom}(f_s)$ if we have $f_s(b_i)<_{\mathbb {N}} f_s(b_j)$ and $f_s(b_i)<_s f_s(b_j)$ but $ f_s(b_i)>_{s+1} f_s(b_j)$ then in $D_{s+1}$ we will remove $b_j$ from $\mathrm {dom}(f_{s+1})$ . Similarly if $f_s(b_i)>_s f_s(b_j)$ but $ f_s(b_i)<_{s+1} f(b_j)$ or $\neg B_s(f_s(b_i), f_s(b_j))$ but $B_{s+1}(f_s(b_i), f_s(b_j))$ .

Next, for each block $b\in \mathrm {dom}(f_s)$ that has not been removed we make sure it has the correct size. Let $y=\min _{\mathbb {N}}\{x: B_s(x,f_s(b))\land \neg B_{s+1}(x,f_{s}(b))\}$ . If y exists, remove points from the end of b so that it has size $|\{x<_{\mathbb {N}} y: B_{s}(x,f_s(b))\}|$ . Now we add points to the end of the block so that the block of $f_{s}(b)$ in $L_{s+1}$ will have the same size as b does in $D_{s+1}$ . Then, in case small numbers have been added we set $f_{s+1}(b)=\min _{\mathbb {N}}\{x: B_{s+1}(x,f_s(b))\}$ .

Then for each block c in $L_{s+1}$ that does not have an element in $\mathrm {range}(f_s)$ we create a corresponding block b in $D_{s+1}$ of the same size as c and set $f_{s+1}(b)=\min _{\mathbb {N}}(c)$ . Finally we densify; for all adjacent $x,y$ which are not part of the same block in $\mathrm {dom}(f_{s+1})$ , we add a new point between x and y. We now have $D_{s+1}$ .

Now the verification. It is clear that D is a computable linear order. We need to make sure it has the right order type. At each stage we densify around the points that are not part of a block, so between adjacent blocks we must have order type $\eta $ .

So we can see that there is an order preserving bijection F from the blocks of D to the blocks of L with $|b|=|F(b)|$ . Hence the order type of D is the same as that of L.

From the construction, if a point is removed from a block then it is never put back in a block at a later stage. So $\mathbf {0}'$ can compute the set of points in D that are not in blocks. As D is computable, $\mathbf {0}'$ can also compute the successor relation on D. From both of these, $\mathbf {0}'$ can compute the block relation.

Proof For the left to right direction we observe that the usual construction (unrelativizing the one given in [Reference Kalimullin, Khoussainov and Melnikov8]) has computable block relation as the blocks that are created are never merged.

For the other direction, since we can compute if two blocks are actually the same block we can make sure we only assign one follower to each block.

Proof of (1) and (2) We will handle the first two statements together. Given a $\Sigma ^0_2$ set A we can assume $A=\mathrm {range}(F)$ for $F(n)=\lim _{s} f(n,s)$ where f is computable. Then we can define a connected approximation of $A=\mathrm {range}(F)$ as follows. Let $\mathrm {dom}(c_n)=\{f(x,n): x< n\}$ and define $c_n(y)=f(x,n+1)$ where x is least such that $f(x,n)=y$ . Clearly $(c_n)_n$ is computable and $\mathrm {range}(c_n)\subseteq \mathrm {dom}(c_{n+1})$ . For each $n,m$ we have that $A_{n,m}=f(B_{n,m},n+m)$ for some $B_{n,m}\subseteq n$ . We take the $B_{n,m}$ which minimizes $\sum _{x\in B_{n,m}}x$ . By construction $\sum _{x\in B_{n,m}}x\ge \sum _{x\in B_{n,m+11}}$ and so the limit $B_{n,\omega }:=\lim _{m} B_{n,m}$ exists. Hence $A_{n,\omega }=F(B_{n,\omega })$ , so we have $(c_n){}_{n}$ is a connected approximation of a subset of A. Consider an $n\in \omega $ . Let $t> n$ be a stage after which $f(m,s)=F(m)$ for all $s\ge t, m\le n$ . Then $F[n]\subseteq A_{t,0}$ , so $F[n]\subseteq A_{t,\omega }$ . So $(c_n)_n$ is a connected approximation of A. Notice that if $f(n,s)$ is monotonic in s then $(c_n)_n$ is also monotonic.

Now consider a computable connected approximation $(c_n)_n$ of a set A. We define a computable function $f:\omega ^2\rightarrow \omega $ as follows. $f(n,0)=0$ , $t_0=0$ . Define $f(n,s+1)$ as follows: $f(n,s+1)=c_s(f(n,s))$ if $n<t_s$ . Let $m_0,\dots , m_{k-1}$ list $\mathrm {range}(c_s)\setminus \mathrm {range}(c_s\circ c_{s-1})$ in order. Define $f(t_s+i,s+1)=m_i$ and $t_{s+1}=t_s +k$ . For $n\ge t_{s+1}$ let $f(n,s+1)=0$ . We have that $A_{n,m}=\{f(x,n+m): x\le t_{n}\}$ and so $\mathrm {range}(F\restriction t_n) = A_{n,\omega }$ and hence $ \mathrm {range}(F)=A$ . Notice that if $(c_n){}_{n}$ is monotonic then F is limitwise monotonic.

Proof Suppose we have a strong $\eta $ -s-representation L of A. We can assume that L has domain $\omega $ and let $L_s=L\restriction s$ . Let $B_s$ be the blocks of $L_s$ according to $S_L$ . For blocks $b,c\in B_s$ and $t\ge s$ we use $|b|_t$ to denote the size the block has in $L_t$ , and we use $b<_t c$ and $b=_t c$ to denote the order of the, possibly merged, blocks in $L_t$ .

We start with $c_0=\emptyset $ , $t_0=0$ . At stage s we assume we are given $\mathrm {dom}(c_s)$ , $t_s$ and a block $b_s\in B_{t_s}$ . We assume that for any $c, d\in B_{t_s}$ with $c<_{t_s}d\le _{t_s} b$ we have $|c|_{t_s}<|d|_{t_s}$ and there are at least s many points between c and d in $L_{t_s}$ . We also assume $\mathrm {dom}(c_s)= \{|b|_{t_s} : b\in B_{t_s}\land b\le _{t_s} b_s\}$ .

We let $b_{s+1}=\max _{<_{t_s}} B_{t_s}$ . Search for a $t>t_s$ such that for every $c,d\in B_t$ with $c<_t d\le _t b_{s+1}$ we have $|c|_t<|d|_t$ and there are at least $s+1$ many points between c and d in $L_{t_s}$ . The fact that L is a strong $\eta $ -s-representation guarantees that we will find such a t. We let $t_{s+1}=t$ and $\mathrm {dom}(c_{s+1})=\{|b|_{t} : b\in B_{t}\land b\le _{t} b_{s+1}\}$ . We define $c_s$ as follows. For $d\le _{t_s} b_s$ we set $c_s(|d|_{t_s})=|d|_t$ . This clearly gives $\mathrm {range}(c_s)\subseteq \mathrm {dom}(c_{s+1})$ . This completes the construction.

Now we need to check that $c_s$ is MOP and meets the condition $\psi $ . If $|d|_{t_s}< |c|_{t_s}$ for $d,c<_{t_s} b_s$ then we have that $d<_{t_s} c$ , so $d\le _t c$ and $|d|_t\le |c|_t$ . Thus $c_s$ preserves $\le $ . Since L is a strong $\eta $ -s-representation, we have that $|d|_n\le |d|_m$ for $n\le m$ , and so $c_s$ is monotonic. If we combine this with the fact that there are s many points between relevant blocks in $B_{t_s}$ we have that if $d_1<_{t_s}\dots <_{t_s} d_n\le _{t_s}b_s$ but $d_1=_t\dots =_t d_n$ then we have $|d_i|_t+s \ge \sum _{i=1}^n (|d_i|_{t_s} +s)$ . So we can conclude that $c_n$ meets the condition $\psi $ .

All that is left to check is that the limits exist and that they give us A. We have that $A_{n,m}=\{|d|_{t_{n+m}}: d\in B_{t_n}, d\le b_n\}$ . Since $B_{t_n}$ is a finite set and each block in $B_{t_n}$ only changes size finitely often, we have that the limit $A_{n,\omega }$ exists and $A_{n,\omega }\subseteq A$ . On the other hand, every $d\in B_L$ is in some $B_n$ , so the there is a stage s such that $t_s>n$ and then we have $|d|_{t_{s+1}}\in \mathrm {dom}(c_{s+1})$ . Thus $|d|\in A_{s+1,\omega }$ . So we have that $A=\cup _n A_{n,\omega }$ and $(c_n)_n$ is a connected approximation of A as desired.

Now for the other direction. Suppose we have a connected approximation $(c_n)_n$ of A satisfying the conditions of the theorem. We construct an $\eta $ -s-representation as follows. The main idea is that at stage s we will have a linear order $L_s$ with successor relation and blocks $B_s$ strictly ordered by size with s many points in between, and the sizes of blocks of $B_s$ are the members of $\mathrm {dom}(c_s)$ .

We define a computable function $H(L,c,m)$ that takes a finite linear order L with successor, a finite function c and a number m, and outputs a finite linear order D with successor extending L if it can. We assume that the blocks $B_L$ are ordered by size the same way they are by $<_L$ . We assume that $\{|b|: b\in B_L\}\subseteq \mathrm {dom}(c)$ . We build D in steps as follows. If $c(|b|)=c(|d|)$ then we merge blocks b and d and all the points in between into one large block. This gives us a $D_0$ that differs from L only in the successor. We then go through each block b of $D_0$ , and if d was a block of L and $d\subseteq b$ then we possibly add points to the end of b so that $|b|=c(|d|)$ . If we have $|b|>c(|d|)$ already then H fails. If H does not fail then this gives us $D_1$ . Now, for each $n\in \mathrm {range}(c) \setminus \{|b|: b\in B_{D_1}\}$ , we add a new block of length n to $D_1$ , keeping the ordering of blocks by size. This gives us a $D_2$ . Finally, between each pair of adjacent blocks in $D_2$ , we add points in a dense way so that there are exactly m many points between them. This is D. If one of the assumptions was wrong then H fails, otherwise it succeeds, and D is a linear order with blocks ordered by size the sizes of which are $\mathrm {range}(c)$ , and there are exactly m many points between adjacent blocks.

We define our strong $\eta $ -s-representation to be $L=\bigcup _s L_s$ where $L_0=\emptyset $ and $L_{s+1}=H(L_s,c_s, s+1)$ . From the definition of H and the fact that each $c_n$ preserves $\le $ and satisfies $\psi $ we can see, using an induction argument, that H will always succeed, so the $L_s$ are all well-defined. From the definition of H we can see that if two points are never part of the same block for some $L_s$ then there is a point in between them. So we have that the successor relation on L is $S_L=\bigcup _n S_{L_n}$ . So L is a computable linear with c.e. successor relation. As the successor relation of a computable linear order is always co-c.e. we have that $S_L$ is computable.

By construction we have that $A_{n,m}=\{|b|_{n+m}: b \text { is a block in } L_{n+1}\}$ , and for every block b in $L_n$ , $c_m(|b|_m)=|b|_{m+1}$ . So we have that $A_{n,\omega }=\{|b|_L: b \text { is a block in } L_n\}$ and L is an $\eta $ -s-representation of A. As the blocks of $L_n$ are ordered by increasing size so too are the blocks of L, so L is a strong $\eta $ -s-representation of A.

Proof The construction goes as follows. We use an enumeration of $L=\{x_i:i\in \omega \}$ , and at stage s we look at the maximal blocks of $L_s$ . We pick rationals to represent the blocks with the idea that $F(r)$ is the size of the block represented by r, but the block that r represents may change when blocks change. To keep track of what blocks rationals follow we will use a sequence of helper functions $h_s:\mathbb {Q}\rightarrow B_{t_s}$ with $\mathrm {dom}(h_s)=\{r: f(r,s)>0\}$ . Once we see a block b appear in $L_s$ it can only grow, so it will remain a block in $L_t$ for $t>s$ . Like we did in the proof of Theorem 3.3 we will use $|b|_t$ , $b=_t c$ and $b<_t c$ to denote the size and order of the blocks from $L_s$ according to $L_t$ . We let $B_t$ be the set of blocks from $L_t$ .

At stage $0$ we start with $f(r,0)=0$ for all $r\in \mathbb {Q}$ and $t_0=0$ . At stage s let $b=\max (B_{t_s})$ . Let $t_{s+1}=t$ be the least stage $t>t_s$ such that for each $c<_t d\le _t b$ in $B_{t}$ we have that $|c|_t <|d|_t$ and there are at least $g(|c|_t+|d|_t)$ many points between c and d in $L_t$ , and furthermore for all n such that $g(n)\le |b|_t+1$ we have $|\{c\in B_t: c\le _t b\land |c|< g(n)\}|\ge n$ . As L is an $\eta $ -s-representation of A there must be such a t.

Let $r_0<\dots <r_{n-1}$ be the domain of $h_s$ ; we begin defining $h_{s+1}$ as follows. Let $h_{s+1}(r_0)$ be the smallest block $c_0\in B_t$ such that $c_0\le _t h_{s}(r_0)\land |c_0|_t\ge f(r_0,s)$ . Let $h_{s+1}(r_i)$ be the smallest block $c_i\in B_t$ such that $h_{s+1}(r_{i-1})<_t c_i\le _t h_{s}(r_i)\land |c_i|_t\ge f(r_i,s)$ .

For each block $c\le _t b$ that is not in $\mathrm {range}(h_{s+1})$ we pick a rational $r_c$ and set $h_{s+1}(r_c)=c$ so that $h_{s+1}$ is order preserving and has image $\{c\in B: c\le _t b\}$ . We define

$$ \begin{align*}f(r,s+1)=\begin{cases} |h_{s+1}(r)|_t, & r\in \mathrm{dom}(h_{s+1}),\\ 0, & \text{otherwise}. \end{cases}\end{align*} $$

Now for the verification: first we need to show that the recursive definition of $h_{s+1}(r_i)$ actually works. Suppose it does not. Then let i be least such that we cannot find a block for $r_i$ . $|h_s(r_i)|_t\ge |h_s(r_i)|_{t_s}\ge f(r_i,s)$ so if $h_s(r_i)$ does not work then there must be some smaller $r_j$ with $h_{s+1}(r_j)=h_s(r_i)$ . So $i>0$ . We have $h_{s+1}(r_{i-1})\le _t h_s(r_{i-1})<_{t_s} h_{s}(r_i)$ , so it must be that $h_s(r_{i-1})$ and $h_{s}(r_i)$ have merged. So $|h_s(r_i)|_t> g(|h_s(r_{i-1})|_{t_s}+ |h_s(r_i)|_{t_s})$ . So by our choice of $t_s$ we have at least $|h_s(r_{i-1})|_{t_s}+ |h_s(r_i)|_{t_s}$ many blocks before $h_s(r_i)$ and at least $|h_s(r_i)|_{t_s}$ have size at least $|h_s(r_{i-1})|_{t_s}$ . But $i\le |h_s(r_i)|_{t_s}$ , so we would have chosen $h_{s+1}(r_{i-1})$ to be one of these, a contradiction.

From the definition of $h_s$ we can see that $h_{s+1}(r)\le _L h_{s}(r)$ for each r and s, so as the blocks of L are well ordered, $\lim _s h_{s}(r)$ exists. From the definition of f we have that it is limitwise monotonic and $F(r)=|\lim _s h_s(r)|_L$ . So $\mathrm {range}(F)\subseteq S$ . If b is a block of L then after some stage t, all of b is in $L_t$ as well as all smaller blocks. So at some stage s, $t_s>t$ , so at stage $s+1$ we have an r such that $h_{s+1}(r)=b$ and for any $n>s$ we have $h_n(r)= b$ as the blocks in that part of the linear order no longer change.

So $S=\mathrm {range}(F)$ as desired.

Proof Suppose that A is a set with a strong $\eta $ -s-representation. Let $(c_n)_n$ be a computable MOP connected approximation of A satisfying condition $\psi $ of Theorem 3.3. We build an connected approximation $(d_n)_n$ of $A\oplus \omega $ satisfying $\psi $ as follows. The first idea is to use $c_m$ with m much larger than n to build $d_n$ . We want m to be large enough that when we see $c_m(x)=c_m(y)$ we can merge the corresponding numbers $2x,2y\in \mathrm {dom}(d_n)$ without violating $\psi $ . The second idea is that when we see $c_m(x)>x$ without any merging, we shift the representative of x in $\mathrm {dom}(d_n)$ to a lager number so that we can handle the case where the gaps between numbers shrink, i.e., when $c_m(y)-c_m(x) < y-x$ for $y>x$ .

To start let $d_0=\emptyset $ and $m_0=0$ . We will ensure that $m_n> \sum _{x\in \mathrm {dom}(d_n)} (x+n)$ , and if $2x\in \mathrm {range}(d_n)$ then $x\in \mathrm {range}(c_{m_n})$ . Given $d_n$ and $m_n$ , let $N>m_n$ be the least number such that if $ x=\max (A_{m_n, N+1})$ then $N> 2x(2x+n+1)$ . Let $m_{n+1}=N$ . This will ensure that $m_{n+1}> \sum _{x\in \mathrm {dom}(d_{n+1})} (x+n+1)$ . There must be such an N as $(c_n)_n$ is a valid connected approximation, so eventually x will stabilize. Let $c=c_{N}\circ \dots \circ c_{m_n+1}$ .

Let $z\in \mathrm {dom} (c)$ be the least such that there is $y>z$ , $c(z)=c(y)$ . For $y\ge 2z\in \mathrm {range} (d_n)$ we define $d_{n+1}(y)= 2c(z)$ . By assumption on $m_n$ , this will not violate condition $\psi $ . If there is no such z then set $z=\max (\mathrm {dom}(c))+1$ . Now we define $d_{n+1}$ on values smaller than $2z$ .

Let $a_0,\dots ,a_{s-1}$ list the elements of $(\mathrm {dom}(c)\oplus \omega )\cap 2z-1$ . Let $b_0,\dots , b_{s-1}$ list the first s elements of $\mathrm {range}(c)\oplus \omega $ . Note that $b_{s-1}<2c(z)$ if $c(z)$ is defined. We define $d_{n+1}(a_i)=b_i$ . This is definitely order preserving, and it is monotonic because c is injective and monotonic on $\mathrm {dom}(c)\cap z$ . This completes the construction of $(d_n)_n$ .

Verification: By construction we can see that $(d_n){}_{n}$ is MOP and satisfies $\psi $ ; all that is left to check is that it is a connected approximation of $A\oplus \omega $ . Let $x\in \mathrm {dom}(d_n)$ and let $2y$ be the least even number in $\mathrm {dom}(d_n)\setminus x$ . Following the construction we can see that $d_n(x)\le d_n(2y)\le 2(c_{m_{n}}\circ \dots \circ c_{m_{n-1}+1})(y)$ . So if we repeat this then we can see that the limit of $x,d_n(x),d_{n+1}(d_n(x)),\dots $ if it exists is less than the limit of $2y,2c_{m_n}(y),2c_{m_n+1}(c_{m_n}(y)),\dots $ , so by monotonicity it exists. Thus $(d_n)_n$ is a valid connected approximation.

Fix x. Let s be a stage at which $c_n\restriction x= c_s\restriction x$ for all $n\ge s$ . Then by the construction we have that $d_n\restriction 2x= d_s \restriction 2x$ for all $n\ge s$ and $\mathrm {dom}(d_n)\cap 2x = (\mathrm {dom}(c_n)\cap x) \oplus x$ . So $(d_n)_n$ is a connected approximation of $A\oplus \omega $ .

Question 5.1. Is there a characterization of the sets with strong $\eta $ -representations that is not in terms of linear orders?

Question 5.2. Is there a set with a $\Delta ^0_2$ strong $\eta $ -s-representation, but no computable strong $\eta $ -s-representation?